Ball from platform with specific vertex On earth in a vacuum. You throw a platon from a platform height 
$h$ and want it to land at point $d$ distant. Note, h is absolutely fixed and d is absolutely fixed. It "must land" at point d, no matter what. You throw it with velocity expressed using $Vx$ and $Vy$.
Now,
(Problem AA)
you want the vertex HEIGHT to be at a percentage $H$ >100 of $h$ (say $H=120 \% $).
-- ~~ ~~ OR ~~ ~~ --
(Situation BB)
you want the vertex DISTANCE to be at a percentage $D$ <50 of $d$ (say $D=25 \% $).
NOTE: the two gentlemen below have generously explained that you CANNOT choose BOTH H and D. Thank you very much for this insight and proof!
So! For each of ProblemAA and ProblemBB, how to calculate $Vx$ and $Vy$ ?
If this is possible - thank you!
{Aside: I assume there's only one $Vx$ / $Vy$ solution for a given value in either ProblmeAA or ProblemBB - but could there be two, or more??}

 A: Let $t$ be the time taken by the ball to reach to the vertex.
So in vertical direction$$v \sin \alpha -gt=0$$
$$t=\frac{v \sin \alpha}{g}$$
In horizontal direction$$\begin{align*}\frac{Dd}{100}&=v \cos \alpha.t\\
&=\frac{v^2 \cos \alpha \sin \alpha}{g}\quad \quad \cdots(1)\end{align*}$$
Again in vertical direction,applying $v^2=u^2+2as$
$$0=(v \sin \alpha)^2-2g ( \frac{Hh}{100}-h )$$
$$(v \sin \alpha)^2=2g ( \frac{Hh}{100}-h )\quad \quad \cdots(2)$$
Dividing $(2)$ and $(1)$ we get
$$\tan \alpha = \frac{2h(H-100)}{Dd}$$
Similarly we can get $v$ also.
$$v_y = \sqrt{2gh ( \frac{H}{100}-1)} \quad \cdots \cdots \text{from}(2)$$
and $$-v_yT+\frac{1}{2}gT^2=h$$
$$T=\frac{v_y+\sqrt{{v_y}^2+2gh}}{g}$$ 
where T= Time of flight
then
$$v_x = \frac{dg}{v_y + \sqrt{v_y^2+2gh}}$$
Also it can be easily proved that if $D=50\%$ then $h=0$. So, $0\%<D\%\leq50\%$. 

Note:If $h<0$(i.e.below the ground) is allowed then $D\%\geq50\%$ is possible but then we some what have to change the notion of $H\%$.

A: Instead of initial speed $v$ and angle $\alpha$, it's easier to work in Cartesian coordinates, with horizontal and vertical components of velocity $v_x$ and $v_y$.
The height of the vertex is $h + v_y^2/2g$, where $g$ is the acceleration due to gravity. To have a height of $h\cdot\frac H{100}$, you need
$$v_y = \sqrt{2gh\left(\frac H{100}-1\right)}.$$
As the horizontal velocity is constant, the ratio of distance in the horizontal direction is the same as the ratio of the corresponding times. The vertical displacement of the projectile is $h + v_yt - \frac12gt^2$. The time to reach the vertex is $v_y/g$, while the total time to drop to height $0$ is $\big(v_y + \sqrt{v_y^2+2gh}\big)/g$. You want their ratio to be $\frac D{100}$, which implies that
$$v_y = \sqrt{2gh\left(\frac{(D/100)^2}{1-2(D/100)}\right)}.$$
If the initial $h$ is fixed, you can't choose both $H$ and $D$ independently; there will be no solution that has both the $H$ you want and the $D$ you want, unless by coincidence they both give the same value of $v_y$.
The horizontal velocity of the projectile doesn't come into this at all. If you pick one of the equations, say you specify $H$ and decide you don't care about $D$, then after you have the solution for $v_y$ you can always choose $v_x$ to make the projectile land at distance $d$. In fact, since the time of flight is $\big(v_y + \sqrt{v_y^2+2gh}\big)/g$, and the projectile must travel a distance $d$ in that time, you can just set
$$v_x = \frac{dg}{v_y + \sqrt{v_y^2+2gh}}.$$
